# What are the elements of the set A? A = {x is in Z: x=(6y-7)/(2y+1), y is in N}

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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In order to determine the elements of the set A, we'll consider the given constraint of the set, that is that all the elements are integer numbers (x integer).

For x to be integer, the denominator (2y+1) has to be the divisor of the numerator (6y-7).

We'll write the division with reminder:

(6y-7)=3(2y+1)-10

We'll divide both sides by (2y+1) and we'l get:

(6y-7)/(2y+1) = 3-10/(2y+1)

But x = (6y-7)/(2y+1) and x integer, so, in order to obtain x integer, (2y+1) has to be a divisor of 10.

We'll write 10 divisors:

D10=+/-2;+/-5;+/-1;+/-10

(2y+1)=1

2y=0,

y=0

x=(6*0-7)/(2*0+1)

x = -7/1

x = -7

(2y+1)=-1

2y=-2

y=-1

x=(6*(-1)-7)/(2*(-1)+1)

x = 13

(2y+1)=2

2y=1

y=1/2 rejected (not integer)

(2y+1)=-2

2y=-3

y=-3/2 rejected (not integer)

(2y+1)=5

2y=4

y=2

x=(6*(2)-7)/(2*(2)+1)

x=5/5

x = 1

(2y+1)=-5

2y=-6

y=-3

x=(6*(-3)-7)/(2*(-3)+1)

x = 5

(2y+1)=10

2y=9

y=9/2 rejected (not integer)

(2y+1)=-10

2y=-11

y=-11/2 rejected (not integer)

A={-7,1,5,13}

neela | High School Teacher | (Level 3) Valedictorian

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A ={x is in Z : x = (6y-7)/(2y+1) , y is in N}

We know N is the set of all natural numbers: 1,2,3,.....

We know Z is the set of all integers (positve and negative): .....4,3,2,1,0, 1,2, 3..... equally saced on the number line.

x = (6y-7)/(2y+1) .

Since y is in N, y = 1,2,3,....

x1 = (6*1-7)/(2*1+1) = -1/3.

x2 = (6*2-7)/(2*2+1) = 5/5 = 1. So x2 belongs to Z.

x3 = (6*3-7)/(2*3+1) = 11/7.

x4 = (6*4-7)/(2*4+1) = 17/9.  So  x4 < 2

x5 = (6*5-7)/(2*5+1) = 23/11. X5 > 2. So Xn does not take the value 2.

xn = (6n-7)/(2n+1).

Lt n--> infinity xn = (6n-7/(2n+1) = (6-1/n)/(2+1/n) = 3.

We see that  for y  > = 2  and for x in n,  the nth term Xn steadily increase from 1 to 3.

So   1 < xn < 3.

Therefore the only integer value  xn takes is x2 = (6*2-7)/(2*2+1) = 1 when y = 2.

Therefore A = A = {x is in Z: x=(6y-7)/(2y+1), y is in N} = {x: x= 1} .